Iron oxide FeO, crystallises in a cubic lattice with a unit cell edge length of 5.0 Å. If density of the FeO in the crystal is 4.0 g cm⁻³, then the number of FeO units present per unit cell is __________. (Nearest integer)
Given: Molar mass of Fe and O is 56 and 16 g mol⁻¹ respectively. N_A=6.0 × 10²³ mol⁻¹

Solution

Chemistry ma’am · 4 steps

Density links the unit cell to the mole: rearrange d = ZM/(a³ N_A) and solve for Z.

1. The density formula. Mass of one unit cell is Z formula units, each of mass M/N_A. Divide by the cell volume a³ to get the density, then rearrange for Z.

   Z = (d × a³ × N_A)/M

2. Get the units right. a = 5.0 Å = 5.0 × 10⁻⁸ cm, so a³ = 1.25 × 10⁻²² cm³. The molar mass of FeO is 56 + 16 = 72 g mol⁻¹.
3. Put in numbers. Z = (4.0 × 1.25 × 10⁻²² × 6.0 × 10²³)/72 = 300/72 = 4.17.
4. So. There are about 4.17 formula units per cell, which rounds to 4.

The common slip: 5. Using the molar mass of Fe alone (56) instead of FeO (72) gives 300/56 = 5.36, which rounds to 5.

The idea to keep. Whenever density, edge length and molar mass are given, use d = ZM/(a³ N_A). Convert Å to cm before cubing, since density is in g cm⁻³.

Answer from the NTA final answer key. Solution by PrepWiser.